<!DOCTYPE html>
<html lang="zh-cn">
<head>
    <title>多元函数的极限与连续性</title>
    <meta charset="utf-8" />
    <link rel="stylesheet" type="text/css" href="../../css/note.css" />
</head>
<body>

<h2>多元函数的极限</h2>

<ol class="theorem">
	设 `alpha, beta ge 0`, `alpha + beta gt 1`, `m, n` 为正整数, 则
	<span class="formula">
		`x^(n alpha) y^(m beta) = o(sqrt(x^(2n)+y^(2m)))`,
    `quad (x,y) to (0,0)`.
	</span>
  特别取 `m = n = 1` 和 `m = n = 2` 得
  <span class="formula">
		`x^alpha y^beta = o(sqrt(x^2+y^2))`,
		`quad x^alpha y^beta = o(root 4(x^4+y^4))`.
  </span>
</ol>

<ol class="proof">
	由不等式 `|x| le root(2n)(x^(2n)+y^(2m))`,
  `|y| le root(2m)(x^(2n)+y^(2m))` 有
  <span class="formula">
    `(|x|^(n alpha) |y|^(m beta))/(sqrt(x^(2n)+y^(2m)))`
    `le (x^(2n)+y^(2m))^(alpha/2 + beta/2 - 1/2)`.
  </span>
	记 `r = sqrt(x^2+y^2)`, 则 `max{|x|, |y|} le r le 1` 时有
  `x^(2n)+y^(2m) le x^2+y^2 = r^2`, 从而上式小于等于 `r^(alpha+beta-1)`.
  令 `r to 0` 即得结论.
</ol>

<ol class="example">
  <li>`x y`, `x^2`, `x^(1+p)` (`p gt 0`) 等函数皆是 `sqrt(x^2+y^2)`
  的高阶无穷小.</li>
  <li>`x y = (x^(3/4) y^(3/4))^(4/3) = o(root 4(x^4+y^4))^(4/3)`
    `= o(root 3(x^4+y^4))`.</li>
</ol>

<script src="../../js/note.js?type=math"></script>
</body>
</html>
